merination notes

BITSAT BITSAT PAPER P APER  09.09.2012 (PT-02)  PHYSICAL CHEMISTRY

ORGANIC CHEMISTRY TEST SYLLABUS Reaction Mechanism : SN1 of R –X, R –O –R & R –OH, E1 in Alcohols, SN2 of R –X, R –O –R & R –OH, Sni, Intramolecular SN2, Sayt Zeff product, Hoffmann product, E1 CB, Comparison of E2/E1CB Physical Properties Properties of Alkyl halide, Alcohol & Ether. Physical SCQ (32) 1. A solution is a mixture of 0.06 M KCl and 0.06 M KI. AgNO 3 solution is being added drop by drop till AgCl starts precipitating (K sp AgCl = 1 x 10 !10 and Ksp AgI = 4 x 10 !16). The concentration of iodide ion at this instant will be nearly equal to : (A) 4.0 x 10 !5M (B) 4 x 10 !8M (C) 2.4 x 10 !8M (D*) 2.4 x 10 !7M

Sol.

When AgC AgCll Sta Starts rts precip precipita itatin ting g

10 !10 [Ag ] = 0.06 +

4 " 10 !16

 –

at that time conc of [S ] =

10 !10

 –

 = 2.4 × 10

7

Sol.

Heat of neutralization of NH 4OH and HCl is (A) 13.7 kcal/mole (B*) < 13 13.7 kcal/mole (C) > 13.7 kcal/mole NH4OH is a weak base. Heat of netralisation < 13.7 kcal.

3.

The solubility of AgCl will be minimum in

2.

(A) 0.001 M AgNO 3

(B) Pure water a a (C*) 0.01 M CaCl 2

(D) Zero

(D) 0.01 M NaCl

 –

Sol.

0.01 M CaCl2 gives maximum Cl  ions. To keep K sp of AgCl constant, decrease in [Ag +] will be maximum

4.

A weak base BOH (0.1 mole) is titrated with with strong acid HCl (0.08 mole) than than the number of H + ion is (Kb for 4 BOH = 10 ) (A*) 24.08 " 20 3 (B) 4 " 10 10 (C) 6.02 " 1013 (D) None  –

 –

BOH

Sol.

 –

+

HCl

 #  #$

t=0 0.1 mole 0.08 mole t= eq eq. 0.02 mo molle – %  as solution is buffer so  –

 –

4

[OH ] = 10  –

 –

[OH ] = 10

%  [H ] = +

4

"

"

BCl + H2O 0.08 mole

0.02 0.08

1  mole 4

10 !14 " 4 10 ! 4

[H+] = 4 " 10 10 %  No. of H + ion = 4 " 10 = 24.08 " 10+13  –

 –

10

" 6.02 " 1023

Page # 1

5.

 –

A certain buffer solution contains X  and HX with their concentrations related as If the value of K  at 25°C for X  is 10 9, pH of the buffer at 25 °C is : (log 2 = 0.3)

[HX] [X! ]

 = 0.2

5.

 –

A certain buffer solution contains X  and HX with their concentrations related as  –

[HX] [X! ]

 = 0.2

 –

If the value of K b at 25°C for X  is 10 9, pH of the buffer at 25 °C is : (log 2 = 0.3)

Sol.

(A*) 5.7 Kb for X = 10  –

(B) 8.3 & Ka for HX = 10

 –

 –

9

The pH of a

 –

(B*) 12

(C) 12.05

C' 2 Kb = & ' = 0.2 1! ' & [OH ] = C' = 0.05 " 0.2 = 0.01  –

7.

& pOH

=2

& pH

(D) 1.95

= 14  – 2 = 12  –

If equal volume of following solutions are mixed, precipitation of Hg 2I2 (Ksp = 2.5 " 10 (A) 10 4 M Hg22)  + 10 111 M I  –

 –

 –

 –

26

) will occur only onl y with :

(B) 10 5 M Hg22)  + 10 10 M I

 –

 –

(C) 10 11 M Hg22)  + 10 7 M I Sol.

(D) 4.3

M  solution of a weak base, if its K b value at 25 °C is 2.5 " 10 3, will be : [Given log 11.18 = 1.05] 20

(A) 11.95 Sol.

(C) 9.7 & pKa = 5

[X! ] = 5 + 0.7 = 5.7 [HX]

pH = pKa + log10

6.

5

 –

 –

(D*) 10 6 M Hg22)  + 10 9 M I

 –

 –

 –

 –

10 6M Hg22+ + 10 9M I  –

 –

10 !6 & IP = 2

 –

2

"

! 24 / 10 !9  , *  = 10  = 1.25 " 10 25 > Ksp - 2 * 8 .   +  –

& precipitation of Hg 2I2 will occur.  –

8.

In which of the following solutions, the degree of dissociation of H 2O is less than 1.8 " 10 7 % at 25 ° C :

Sol.

(A) 10 6 M HCl (B) 10 7 M NaOH (C) 10 8 M HCl (D*) All of these 7 The degree of dissociation of pure water at 25 °C = 1.8 " 10 % & any H + or OH  ions from an external source will suppress the dissociation of H 2O.

 –

 –

 –

 –

 –

9.

Sol.

10.

Three sparingly soluble salts M2X, MX and MX 3 have their solubility product product in the ratio of 4: 1 : 27. Their solubilities will be in the order : (A) MX3 > MX > M2 X (B*) MX3 > M2X > MX (C) MX > MX 3 > M2X (D) MX > M2X > MX 3 3 1/3 For M2X , 4S1  = 4x ; S1 = x 2 For MX , 4S2  = x ; S2 = x1/2 4 For MX3 , 27S3  = 27x ; S3 = x1/4 S3 > S 1 > S 2 & Calculate the pH of a 0.1 M K 3PO4 solution. The third dissociation constant of phosphoric acid is 10 Given (0.41)1/2 = 0.64 ; log 3 = 0.48  (A  (A) 12.5 (B*) 12.44 (C) 12.25 (D) 12

 –

12

.

Page # 2

Sol.

10 !14 Kw Kh =  =  = 10 K a3 10 !12

 –

2

%

Ch 2 Kh = (1 ! h)

5.

 –

A certain buffer solution contains X  and HX with their concentrations related as  –

[HX] [X! ]

 = 0.2

 –

If the value of K b at 25°C for X  is 10 9, pH of the buffer at 25 °C is : (log 2 = 0.3)

Sol.

(A*) 5.7 Kb for X = 10  –

(B) 8.3 & Ka for HX = 10

 –

 –

9

The pH of a

 –

(B*) 12

(C) 12.05

C' 2 Kb = & ' = 0.2 1! ' & [OH ] = C' = 0.05 " 0.2 = 0.01  –

7.

& pOH

=2

& pH

(D) 1.95

= 14  – 2 = 12  –

If equal volume of following solutions are mixed, precipitation of Hg 2I2 (Ksp = 2.5 " 10 (A) 10 4 M Hg22)  + 10 111 M I  –

 –

 –

 –

26

) will occur only onl y with :

(B) 10 5 M Hg22)  + 10 10 M I

 –

 –

(C) 10 11 M Hg22)  + 10 7 M I Sol.

(D) 4.3

M  solution of a weak base, if its K b value at 25 °C is 2.5 " 10 3, will be : [Given log 11.18 = 1.05] 20

(A) 11.95 Sol.

(C) 9.7 & pKa = 5

[X! ] = 5 + 0.7 = 5.7 [HX]

pH = pKa + log10

6.

5

 –

 –

(D*) 10 6 M Hg22)  + 10 9 M I

 –

 –

 –

 –

10 6M Hg22+ + 10 9M I  –

 –

10 !6 & IP = 2

 –

2

"

! 24 / 10 !9  , *  = 10  = 1.25 " 10 25 > Ksp - 2 * 8 .   +  –

& precipitation of Hg 2I2 will occur.  –

8.

In which of the following solutions, the degree of dissociation of H 2O is less than 1.8 " 10 7 % at 25 ° C :

Sol.

(A) 10 6 M HCl (B) 10 7 M NaOH (C) 10 8 M HCl (D*) All of these 7 The degree of dissociation of pure water at 25 °C = 1.8 " 10 % & any H + or OH  ions from an external source will suppress the dissociation of H 2O.

 –

 –

 –

 –

 –

9.

Sol.

10.

Three sparingly soluble salts M2X, MX and MX 3 have their solubility product product in the ratio of 4: 1 : 27. Their solubilities will be in the order : (A) MX3 > MX > M2 X (B*) MX3 > M2X > MX (C) MX > MX 3 > M2X (D) MX > M2X > MX 3 3 1/3 For M2X , 4S1  = 4x ; S1 = x 2 For MX , 4S2  = x ; S2 = x1/2 4 For MX3 , 27S3  = 27x ; S3 = x1/4 S3 > S 1 > S 2 & Calculate the pH of a 0.1 M K 3PO4 solution. The third dissociation constant of phosphoric acid is 10 Given (0.41)1/2 = 0.64 ; log 3 = 0.48  (A  (A) 12.5 (B*) 12.44 (C) 12.25 (D) 12

 –

12

.

Page # 2

Sol.

10 !14 Kw Kh =  =  = 10 K a3 10 !12

 –

2

%

Ch 2 Kh = (1 ! h)

Sol.

10 !14 Kw Kh =  =  = 10 K a3 10 !12

 –

Kh

as 1  – h = 1, h =

C

as h > 0.1 %  1  – h

0

2

Ch 2 Kh = (1 ! h)

%

10 !2  = 0.316 0.1

 =  1

10 !1 " h 2  or 0.1 (1  – h) = h2 &  10  = (1 ! h)  –

2

or, 0.1  – 0.1 h = h2 or, h + 0.1 h  – 0.1 = 0

! 0.1 ) (0.1)2 ) 4 " 0.1  = 0.27 2

or, h =

as, PO43  + H2O HPO42  + OH   c(1  – h)   ch ch ch &  [OH ] = ch  = 0.1 " 0.27  = 27 " 10 3 pOH = 3  – log 27 = 3 log33 = 3  – 3 log 3 = 3  – 3 " 0.48 = 1.56 pH = 14  – 1.56 = 12.44  –

 –

 –

 –

 –

11.

The pKa of HCN is 9.3. The T he pH of a solution prepared by mixture 2.5 mole of KCN and 2.5 mole of HCN in water and making up the total volume to 500 ml is (A*) 9.3 (B) 7.3 (C) 10.3 (D) 8.3

Sol.

pH = pKa + log [ Acid]  = 9.3

12.

Calculate the molar solubility of AgCl in 2.5 M NH 3 solution. [Given : KspAgCl = 10

[Salt]

(A*) 0.025 mol/L Sol.

(B) 0.2 L mol

K sp

(D) None of these

.... (1)

s

s2

Kf = [ Ag ) ](2.5 ! 2s)2

Sol.

1

[Ag(NH3)2]+

s

13.

(C) 0.4 L mol

 –

Ag+ + 2NH3 

or,

 –

1

 , Kf[Ag(NH3)2]+ = 106]

10

Ag+(aq) + Cl  (aq)

AgCl(s)

[Ag+] =

 –

 –

s  = 10 2 . 5 ! 2s

or, Kf = K (2.5 ! 2s)2 sp

 –

12

or, Kf " Ksp =

s2 2

( 2. 5 ! 2s )

 –

or, 10 2 =

or, s = 0.025  – 0.02 s or, 1.02 s = 0.025 or, s =

0.025 1.02

1

s 2.5 ! 2s

0.025 mol/L

What is the concentration of acetic acid which can be added to 0.5 M formic acid so that the % dissociation of neither acid is changed by the addition. K a for acetic acid is 1.85 × 10-5, Ka for formic acid = 2.4 × 10-4. (A) Any concentration (B) There can not be any concentration (C*) 6.66 M (D) 3.33 M C1 '1 = C2 '2 K a1 C1  =

K a2 C 2

 –

 –

1.8 × 10 5 × C1 = 2.4 × 10 4 × 0.5 C1 = 6.66 M

Page # 3

14.

CH3NH2 (Kb = 5 × 10 4) 0.1 mole of CH 3NH2 (Kb = 5 × 10 4) is mixed with 0.08 mole of HCl and diluted to one litre. What will be the H + concentration in the solution? is mixed with 0.08 mole of HCl and diluted to one litre. What will be the H + concentration in the solution?  –

 –

14.

CH3NH2 (Kb = 5 × 10 4) 0.1 mole of CH 3NH2 (Kb = 5 × 10 4) is mixed with 0.08 mole of HCl and diluted to one litre. What will be the H + concentration in the solution? is mixed with 0.08 mole of HCl and diluted to one litre. What will be the H + concentration in the solution?  –

 (A) 8 × 10 2 M  –

Sol.

 –

11 M

 –

(B*) 8 × 10 2

11 M

 –

(C) 1.6 × 10

(D) 8 × 10 5 M  –

3

CH3NH2 + HCl  #   #$ CH3 NH3 C l .1 .08 .08 .02 0 .08  –

For buffer sol. |OH | = Kb x  –

|OH | =

so

|H+|

 –

2

5  x 10 4

 –

[Base] .02  = 5 x 10 4 x [Salt] .08  –

4

10 !14 4 = 4  = x 10 5 x10 ! 4 5

15.

10

Sol.

(A*) 4 Initially pH = 7 finally [NaOH] = 10

 –

8 x 10

1 M 11

Ans.

mole of NaOH was added to 10 litre of water. The pH will change by (B) 3

 –

So,

10 =

 –

4(pH)

3

(C) 11

(D) 7

so pOH = 3 pH = 11

=4  –

16.

The sum of negative logarithm of hydrogen ion and hydroxide ion concentration at 37 ºC : [Kw = 2.5 × 10 4]  (A) 14 (B*) Less than 14 (C) greater than 14 (D) Data insufficient.

Sol.

pH + pOH = pKW = 15  – log (5)2 = 15  – 2 × .699 = 13.6

17.

In the reaction : [Ag(CN)2] + Zn  #   #$  the complex formed will be : (A*) Tetrahedral (B) square planar (C) octahedral (D) triangal bipyramidal

Sol.

 #$  [Zn(CN) ]2  + 2 Ag 2 [Ag(CN)2]  + Zn  #  4

 –

 –

 –

Tetrahedral

Sol.

All the following complexes show a decreases in their weights when placed in a magnetic balance. Then which of the these has square planar geometry : (A) Ni(CO)4 (B*) K[AgF4] (C) Na2[Zn(CN)4] (D) None of these 8 K [AgF4] is square planar because Ag(555) is 4d  and complex is diamagnetic.

19.

It is an experiment fact that :

18.

DMG + Ni(55)salt + NH 4OH  #   #$ Red ppt. Which of the following is wrong about this red ppt : (A) It is a non –ionic complex (B) It involves intra molecular H –bonding 3 (C*) Ni(55) is sp  hybridised (D) It is a diamagnetic complex Sol.

The complex is

Page # 4

20.

Sodium nitroprusside is a diamagnetic substance and a important laboratory reagent for the testing of sulphide ions. The metal involved in the complexation in this is present in which of the following hybridisation state :

20.

Sodium nitroprusside is a diamagnetic substance and a important laboratory reagent for the testing of sulphide ions. The metal involved in the complexation in this is present in which of the following hybridisation state : (A) sp3 (B) dsp2 (C*) d2sp3 (D) sp3d2

Sol.

Sodium nitroprusside is Na2 [Fe(CN)5 (N O) ] ; a diamagnetic complex.

21.

All the following complex ions are found to be paramagnetic : P : [FeF6]3 ; Q : [CoF6]3 R : [V(H 2O)6]3+ ; S : [Ti(H2O)6]3+ The correct order of their paramagnetic moment (spin only) is : (A*) P > Q > R > S (B) P < Q < R < S (C) P = Q = R = S (D) P > R > Q > S On the basis of number of electrons the correct order is P > Q > R > S. When the complex K 6 [(CN)5 Co –O –O –Co(CN)5] is oxidised by bromine into K5[(CN)5 Co –O –O –Co(CN)5]. Then which of the following statements will be true about this change: (A) Co(55) is oxidised in Co(555) (B) The O –O bond length will increase (C*) The O –O bond length will decrease (D)  A ‘ ’ & ‘B’ both are correct

)

 –

Sol. 22.

 –

 –

 –

Sol.

In the first complex ligand is O 22  which is oxidised into O21  . hence O  – O bond length decreases.

23.

The octahedral complex [Rh(NO 2) (SCN) (en) 2]+ can exist in a total number of isomeric forms including stereoisomers : (A) 2 (B) 4 (C) 8 (D*) 12

Sol.

(1) NO2 / SCN (2) ONO / SCN (3) NO2 / NCS (4) ONO NCS 24.

(5) NO 2 / SCN (6) ONO / SCN (7) NO 2 / NCS (8) ONO / NCS

For the reaction Ni2+ + 4NH3 

(9) (10) (11) (12)

[Ni(NH3)4]2+  –

at equilibrium, if the solution contains 1.6 × 10 4% of nickel in the free state, And the concentration of NH 3 at equilibrium is 0.5 M. Then the instability constant of the complex will be approximately equal to : (A) 1.0 × 10 5 (B) 1.5 × 10 16 (C*) 1.0 × 10 7 (D) 1.5 × 10 17  –

Sol.

 –

Ni2+ + 4 NH3  &

But

or

&

k=

 –

[Ni(NH3)4]2+ [Ni (NH3 ) 4 ] 2) [Ni 2) ] [NH3 ] 4

[Ni2) ] [Ni2) ] ) [Ni (NH3 ) 4 ] 2) Ni 2) [Ni (NH3 ) 4 ] 2 ) k=

 –

1 1.6

10 6 1.6 " (0.5)4

 –

 = 1.6 × 10

6

 –

× 10

6

 = 107

Hence instability constant = 10

 –

7

Page # 5

25.

6 0 In which of the following complex ion, the metal ion will have t 2g , e g  configuration according to CFT::

25.

6 0 In which of the following complex ion, the metal ion will have t 2g , e g  configuration according to CFT::

Sol.

(A) [FeF6]3 (B) [Fe(CN)6]3 (C*) [Fe(CN)6]4 In [Fe(CN)6]4  ; Fe(55) is t2g6 , eg0 due to strong ligands.  –

 –

 –

(D) None of these

 –

26.

Spin only magnetic moment of a complex having CFSE =  – 0.6 40 and surrounded by weak field ligands can be (A) 1.73 BM (B) 4.9 BM (C*) both (A) & (B) (D) None of these

Sol.

The options can give CFSE =  – 0.6 40 with weak field ligands %

27.

Sol.

Which of the following statements is not correct? (a) [Ni(H2O)6]2+ and [Ni(NH3)6]2+ have same value of CFSE (b) [Ni(H2O)6]2+ and [Ni(NH3)6]2+ have same value of magnetic moment (A*) Only a (B) Only b (C) Both a and b Ammonia is a stronger field ligand than water.

28.

The correct IUPAC name of the complex:

d4 and d9.

(D) None of these

OH

H3C

 C = N

•     

•     

CoCl

    •

2

is :

    •

 C = N H3C

OH

(A*) Dichlorodimethylglyoximatecobalt (II) (C) Dimethylglyoximecobalt(II) chloride

(B) Bis(dimethyglyoxime)dichlorocobalt (II) (D) Dichlorodimethylglyoxime-N, N-cobalt (II)

29.

Which of the following pair of complexes have the same EAN of the central metal atoms/ions? (A) [Cu(NH3)4]SO4 and K3[Fe(CN)6] (B) K4[Fe(CN)6] and [Co(NH3)6]Cl3 (C) K3[Cr(C2O4)3] and [Ni(CO)4] (D*) all of the above

Sol.

(A) [Cu(NH3)4]2+ = 29  – 2 + 8 = 35   [Fe(CN)6]3  = 26  – 3 + 12 = 35 (B) [Fe(CN)6]4 = 26  – 2 + 12 = 36 [Co(NH3)6]3+ = 27  – 3 + 12 = 36 (C) [Cr(C2O4)3]3  = 27  – 3 + 12 = 36   [Ni(CO)4] = 28 + 8 = 36  –

 –

 –

30.

In the reaction [CoCl2(NH3)4]+ + Cl

 –

 #   #$  [CoCl3(NH3)3]

+ NH3 only one isomer of product is obtained .

Hence the initial complex must be Sol.

(A) cis isomer Moderate

(B*) trans isomer

(C) both

(D) mixture of both

! Cl #   #  # $

symmetrical

only single product

Page # 6

Cl!  #  #  # $

two isomers product

replacable positions Aq.  #  #  # $

31.

AgNO 3

(A)

' X'  product :

(Major )

(B)

Sol.

Aqueous AgNO3 catalyse SN1 reaction.

32.

Consider the following reaction.

(C*)

(D)

SOCl

2  #   # #   #  $

ether In the above reaction which phenomenon will take place :

(A) Inversion

(B*) Retention

(C) Racemisation

Sol.

It is SNi reaction so retention takes place

33.

Which one of the following has maximum nucleophilicity ?

Sol.

(A*) CH3 S 3 Nucleophilicity 6 size

(B) (in a group).

(D) Isomerisation

(C) Et3N

(D)

O || (C) ! S ! O CH3 || O

O || (D*) ! O ! S ! CH3 || O

NaCN

34.

 #   # #   #  $

In the given reaction rate is fastest, when (X) is :

(A)  –OH

(B) –NH2

Sol.

Leaving group ability 6 Stability of anion.

35.

In the following reaction the most probable product will be :

Page # 7

(A)

(B*)

(C)

(D)

(A)

(B*)

Sol.

36.

Sol.

(C)

(D)

7

When the concentration of alkyl halide is tripled and the concentration of of SN2 reaction increases by: (A) 3 times (B) 2 times (C*) 1.5 times Rate of SN2 ' [R  – X ] [Nu ]

ion is reduced to half, the rate (D) 6 times

 –

[3RX] 21 OH – r2 r1  = [RX] [OH – ]

r2 = 1.5 r1 37.

Sol.

In which of the following reaction the product obtained is t-butyl methyl ether ?

conc.H2SO4 (A) CH3OH + HO —CH2—CH3  #   # #   #   #   #   #  $

(B)

(C*)

(D)

t-butyl methyl ehter is a mixed ether and for the preparation of mixed ethers in high yield the essential condition is the use of primary alkyl halide. Thus,

 #  # $ !NaBr

This reaction is williamson's synthesis. 38.

Sol.

PCl

alc KOH

5$  # #   #   # $  B CH3CH2CH2OH  #  A  #  # #  #  B is identified as : (A) propanal (B) propane

(C) propyne

(D*) propene

PCl5 or SOCl2 or

 # #   #   #   #   #   #  $ RCl ROH  # 

PX 3 ( X :Cl, Br, I) or ZnCl2 / HCl PCl

5$  # #  CH3 CH : CH2 CH3CH2CH2OH  #  !

HCl

Alkene

B is an alkene (propene) Page # 8

39. Sol.

The only alcohol that cannot be prepared by the indirect hydration of alkene is : (A) ethyl alcohol (B) propyl alcohol (C) isobutyl alcohol (D*) methyl alcohol Methyl alcohol cannot be prepared by hydration of alkene as simplest alkene has two carbons so alcohol of

39. Sol.

The only alcohol that cannot be prepared by the indirect hydration of alkene is : (A) ethyl alcohol (B) propyl alcohol (C) isobutyl alcohol (D*) methyl alcohol Methyl alcohol cannot be prepared by hydration of alkene as simplest alkene has two carbons so alcohol of at least two carbon atoms can be formed. H3 C | (H2SO4 ) H2C ) H)

 # $ H2C = CH2  #  #

HSO!4  #   # #   #  $

H3C HOH |  #  # # $ H2SO4 ) CHCH2OH CH2OSO 3H Ethanol H2C = CH  – CH3 HOH

 #  # # $ H2SO 4 (Markowniko ff 's rule)

40. Sol.

H3C ! CH ! CH3 | OH

) HO ! CH 2 ! CH 2 CH3

Minor product

Major product

Lucas reagent reacts fastest with : (A) butanol –1 (B) butanol –2 (C*) 2 –methyl –propanol –2 The order of reactivity with alcohols with lucas reagent is 3º > 2º > 1º & Lucas reagent reacts fastest with 3 º alcohol.

(D) 2 –methyl –propanol –1

(a)

(b)

(c)

(d) & choice

(C) is the answer as it is 3 º alcohol and rate of reaction is f astest for 3º alcohol.

Page # 9

 BITSAT- PT - 2 - XII - (09-09-12) Syllabus : Sequence&Series, P&C, Binomial Theorem, M athematical Induction, Determinant, Straight lines

 BITSAT- PT - 2 - XII - (09-09-12) Syllabus : Sequence&Series, P&C, Binomial Theorem, M athematical Induction, Determinant, Straight lines 1. Sol.

2. Sol.

3. Sol.

For every natural number n, n(n + 3) is always (A*) even (B) odd (C) multiple of 4 Let P(n) = n(n + 3), then P(1) = 1(4) = 4 which is even and multiple of 4. P(2) = 2(5) = 10 which is even and multiple of 5. P(3) = 3(6) = 18 which is even. Hence it is clear that P(n) is even ! n " N The greatest positive integer which divides 3 2n  – 2n  – 1 ! n " N is (A) 1 (B*) 2 (C) 4 2n Let P(n) = 3  – 2n  – 1, then P(1) = 3 2  – 2  – 1 = 6 P(2) = 3 4  – 4  – 1 = 76 P(3) = 3 6  – 6  – 1 = 722 Obviously 2 is the greatest positive integer which divides P(n) ! n " N.

Sol.

a % 28d (a % 8d) % 20d 20d 2 = =  = a % 18d (a % 8d) % 10d 10d 1

The solution of the equation (x + 1) + (x + 4) + (x + 7) + ....+ (x + 28) = 155 is (A*) 1 (B) 2 (C) 3 (D) 4 We have (x + 1) + (x + 4) + (x + 7) + ....+ (x + 28) = 155 Let n be the number of terms in the A.P. on L.H.S. Then x + 28 = (x + 1) + (n  – 1)3 $ n = 10 & (x + 1) + (x + 4) +.....+ (x + 28) = 155

$ 5.

10 [(x + 1) + (x + 28)] = 155 $ x = 1 2

If the arithmetic and geometric means of a and b be A and G respectively, then the value of A  – G will be

Arithmetic mean of a and b = A = and geometric mean G = a % b Then A  – G =  – 2

6. Sol.

, a  – b ) ' (C*) * 2 '( +*

a % b (B) 2

a  – b (A) a

Sol.

(D) 8

If the 9th terms of an A.P. be zero, then the ratio of its 29th and 19th term is (A) 1 : 2 (B*) 2 : 1 (C) 1 : 3 (D) 3 : 1 Given that 9th term = a + (9  – 1) d = 0 $ a + 8d = 0 Now ratio of 29th and 19th terms =

4.

(D) multiple of 5

2

(D)

2ab a%b

a % b 2

ab

a % b – 2 ab ( a ) % ( b )  – 2 ( a )( b ) , a  – b ) '  = = * ab = 2 2 2 *+ '( 2

2

In series 1,2,2,2,2,3,3,3,3,3,3,3,3,3, 4,...........the 400th term is (A) 9 (B) 10 (C*) 11

-

Number 11 starts at (12 % 22 % ..... % 10 2 ) % 1

.

th

2

(D) 12

position

i.e. 386th position.

7.

The sum of the series

(A)

n (n2 % 1)

1 2

1% 1 % 1 (B*)

n2 % n % 1

4

%

2 2

1% 2 % 2

n (n % 1) 2

2(n % n % 1)

4

%

3 2

1% 3 % 34 (C)

% .........  to n terms is

n (n2 % 1) 2(n 2 % n % 1)

(D)

n2 % n 2 Page # 1

Sol.

Let Tn be the n th term of the series

1 2

4

1% 1 % 1

%

2 2

1% 2 % 2

4

%

3 2

1% 3 % 34

% .........

Sol.

Let Tn be the n th term of the series

1 1 % 12 % 14 Then Tn =

%

Now

1 % n2 % n4

/

Tr 0

r 01

Sol.

=

3 1 % 32 % 3 4

% .........

n 2 2

(1 % n )  – n

2

 =

n 2

2

(n % n % 1)(n  – n % 1)

 =

1,

1 1 )  – * 2 + n 2  – n % 1 n 2 % n % 1'(

) 1, 1 1 1 ) 1 , 1 1 ) 1 , 1 1 ) 1 ,1  –  –  –  – ' + * + * +...+ * * ' ' ' 2 +1 % (n – 1) n 1 % n(n % 1) ( 2 +1 1 % 1.2 ( 2 +1 % 1 .2 1 % 2.3 ( 2 +1 % 2.3 1 % 3.4 (

) 1, 1 n (n % 1) 1 – * '  = 2 + 1 % n (n % 1) ( 2(n 2 % n % 1)

=

9.

1 % 22 % 2 4 n

n

Sol.

%

) 1, 1 1  – * % ' 2 +1 (n – 1) n 1 % n(n % 1) (

=

8.

2

The number of common terms to the sequence 17, 21, 25, .......417 and 16, 21, 26,.....466 is (A) 21 (B) 19 (C*) 20 (D) 22 Common terms are 21,41,61,......(d = LCM of 4,5 = 20) tn 1 417 $ 21 + (n  – 1)20 1 417 $ n 1 20.8 $ max value of n = 20 If a, b, c are in A.P. as well as in G.P., then (A) a = b 2 c (B) a 2 b = c a % c 2

As given b =

(C) a 2 b 2 c

(D*) a = b = c

......(i)

and b2 = ac $ (a + c)2 = 4ac $ (a  – c)2 = 0 $ a = c putting a = c in (i) we get b = c & a =b =c 1

10.

If G be the geometric mean of x and y, then

(A) G2 Sol.

(B*)

As given G = 1

&

2

G  – x

Sol.

%

G  – x

(C)

G2

2

%

1 G  – y 2

2 G2

2

 =

(D) 3G2

xy 1 2

G  – y

1 2

=

xy  – x

1

 +

2

xy  – y 2

8  1 1 5 1 1 66 – % 33 =  = G2 7  x y 4 xy

1 = x  – y 11.

2

1

2

If a 1, a 2, a 3,........ are in A.P. such that a 1 + a 5 + a 10 + a 15  + a 20 + a 24 = 225, then a1 + a 2 + a 3 + ..... + a 23  + a 24 is equal to (A) 909 (B) 75 (C) 750 (D*) 900 a1 + a 5 + a 10 + a 15  + a 20  + a 24 = 225

$

3 (a 1 + a 24) = 225

(sum of terms equidistant

from beginning and end are equal) a 1 + a 24 = 75 Now =

a1 + a 2 + ........ + a 23  + a 24

24  [a 1 + a 24] = 12 × 75 = 900 2 Page # 2

12.

There are n distinct points on the circumference of a circle. The number of pentagons that can be formed with these points as vertices is equal to the number of possible triangles. Then the value of n is(A) 7 (B*) 8 (C) 15 (D) 30

Sol.

n

C

n

C

12.

There are n distinct points on the circumference of a circle. The number of pentagons that can be formed with these points as vertices is equal to the number of possible triangles. Then the value of n is(A) 7 (B*) 8 (C) 15 (D) 30

Sol.

n

C5  = n C 3

n

C5  = n Cn – 3

5 = n  – 3 n=8 13.

Sol.

14.

Sol. 15.

Sol.

A question paper consists of two parts A and B. Part A has 4 questions in which each question has an alternative and part B has 3 questions without any alternative. The number of ways to attempt paper when at least one question must be attempted for each part is (are) (A) 561 (B*) 560 (C) 648 (D) 127 4 3 Required ways = (3  –1)(2  – 1) = 560

Number of ways such that 6 boys and 3 girls can be seated such that there is exactly one boy in between any two girls (A) 50400 (B*) 21600 (C) 10800 (D) 36000 Number of ways = 6! × 5 × 3! All letters of the word 'RACHIT' are permuted in all possible ways and the words so formed (with or without meaning) are written as in dictionary, then the 484 th word is(A) RACHIT (B*) RACITH (C) RACTHI (D) RACIHT $ ACHIRT 5! CAHIRT 5! $ $ HACIRT 5! IACHRT 5! $ th RACHIT is 481  word RACHIT th RACHTI is 482  word RACIHI is 483th word RACITH is 484th word

16.

The number of ways in which 6 different red roses and 3 different white roses can form a garland so that all the white roses come together is (A*) 2160 (B) 2165 (C) 2155 (D) 4320

Sol.

ways =

17.

There are 10 points in a plane of which no three points are collinear and 4 points are concyclic. The number of different circles that can be drawn through at least 3 of these points is (A) 116 (B*) 117 (C) 120 (D) 115 10 4 Total number of solutions = C3  – C3 + 1 = 117.

Sol.

18.

(7 – 1)! 9 3! = 2160 2

How many different arrangements can be made out of the letters in the expansion A 2B3C4, when written in full length ? 9!

(A*) 2! 3 ! 4! Sol.

Sol.

9! 2. 3 . 4

(C) 2 ! 3 !  – 4

9!

(D) 2! % 3 ! % 4 !

Here A, B, C are repeated twice, thrice and four times respectively

& 19.

(B)

No. of arrangements =

(2 % 3 % 4)! 9!  = 2! 3 ! 4 ! 2! 3 ! 4 !

Number of positive integral solutions of x 1. x2 x3 = 210 is(A) 25 (B) 26 (C) 27 (D*) 81 We have x1 x2 x3 = 210 = 2.3.5.7 Total no. of solutions of the equation x 1x2x3 = 210 is 3 × 3 × 3 × 3 = 81 & Page # 3

20.

Sol.

Total number of ways in which 15 identical blankets can be distributed among 4 persons so that each of them gets at least two blankets, equal to (A*) 10C3 (B) 9C3 (C) 11C3 (D) none of these Let 4 persons recieve B1, B2 , B3, B4 number of blankets

20.

Sol.

21.

Sol.

22.

Sol.

Total number of ways in which 15 identical blankets can be distributed among 4 persons so that each of them gets at least two blankets, equal to (A*) 10C3 (B) 9C3 (C) 11C3 (D) none of these Let 4 persons recieve B1, B2 , B3, B4 number of blankets {B1,B2 , B3 ,B4 : 2} & B1 + B2 + B3 + B4 = 15 Thus number of ways = 10C3 Let S (k) = 1 + 3 + 5 +.......+ (2k  – 1) = 3 + k 2. Then which of the following is true ? (A) S(1) is correct (B*) S(k) $ S (k + 1) (C) S(k) S(k + 1) (D) Principle of mathematical induction can be used to prove the formula S(k) = 1 + 3 + 5 + ..... + (2k  – 1) = 3 + k2 put k = 1 in both sides, we get LHS = 1 and RHS = 3 + 1 = 4 & $ LHS 2 RHS Put (k + 1) in both sides in the place of k LHS = 1 + 3 + 5 + .... + (2k  – 1) + (2k + 1) RHS = 3 + (k + 1)2 = 3 + k 2 + 2k + 1 Let LHS = RHS 1 + 3 + 5 + .......... + (2k  – 1) + (2k + 1) = 3 + k 2 + 2k + 1 $ 1 + 3 + 5 + ...... + (2k  – 1) = 3 + k2 If S(k) is true, then S(k + 1) is also true. $ Hence, S(k) S(k + 1). If the coefficients of second, third and fourth terms in the expansion of (1 + x) 2n are in A.P., then which of the following is TRUE. (A) n2  – 9n + 7 = 0 (B) 3n2  – 9n + 7 = 0 (C) 3n2 + 9n + 7 = 0 (D*) 2n 2  – 9n + 7 = 0 2n C1 , 2nC2 , 2nC3 are in A.P. $ 2n2  – 9n + 7 = 0 9

23.

8 3 2 1  5 3  is The term that is independent of x in the expansion of 6 x ; 3 x 4 7 2 5

8 3 5 8  1 5 (A) C 6 6 3 6 ; 3 7 2 4 7  3 4

4

9

9

Sol.

9

8 3 2 1  5 6 x ; 3  = 3 x 4 7 2

/

8 1 5 (B*) C3 6 3 7 6 4

3

4

8 3 5 8  1 5 (C) C 4 6 3 6 ; 3 7 2 4 7  3 4

9

9

r 00

8 3  5 Cr 6 x 2 3 7 2  4

9 ;r

8  1  5 6 – 3 7  3 x 4

5

9

6

8 3 5 8  13 5 3 (D) C 6 6 3 6 ; 7 2 4 7  3  4

6

9

r

For the term that is independent of x 18  – 2r  – r = 0 $ r = 6 3

6

8 3 5 8  1 5 8 1 5 Required term = C 6 6 3 6 ; 3  = 9 C 6 6 3 7 2 4 7  3 4 7 6 4

3

9

24.

In the expansion of (2 1/5 +

20 3 ) , the sum of all rational terms is equal to (B) 84 (C) 97 (D*) none of these

(A) 21 Sol.

Tr+1 = =

20

20

Cr (21/5)20 r  ( 3 )r

Cr . 2

 –

4;

r 5

r

.32

As 2 and 3 are relatively primes. T r+1 is rational, if

r r  and  are integer ’s 5 2

& r is multiple of 10 & 0 1 r 1 20 r = 0 ,10, 20 Thus sum of rational terms = T1 + T11 + T21 = 20C0 24 + 20C10 22. 35 + 20C20 . 310 This is more than 21, 84, 97 Page # 4

25.

B< 3 2003 ?< The value of A 28 >  where {.} denote the fractional part, is equal to 0

Sol.

38. Sol.

Since 3.3  – 4.4  – 8 =  – 15 < 0

$

3x  – 4y  – 8 > 0 $

$

5y <  – 8 $

8  y  5 3 7  3 4

3 6 –

y <  –

 –

 4y  – 8 > 0

8 5

If the lines ax + 12y + 1 = 0, bx + 13y + 1 = 0 and cx + 14y + 1 = 0 are concurrent, then a, b, c are in(A) H.P. (B) G.P. (C*) A.P. (D) None of these Since the given lines are concurrent, a 12 1 b 13 1

&

a – b  –1 0 b

$

 = 0

c 14 1

1 2 (B) 3 3 Equation of the line through the point (2, 2)

and H to line (1) is (y  – 2) =

$ $

Sol.

4 3

4 . [Putting x = 0] 3

The line segment joining the points (1, 2) and (k, 1) is divided by the line 3x + 4y 4 : 9, then k is(A*)  – 2 (B) 2 (C)  –3 (D) 3 L : 3x + 4y  – 7 = 0  –  L (1, 2) : L (k, 1) = 4 : 9 $ –  (3 + 8  – 7) : (3k + 4  – 7) = 4 : 9 k =  – 2. $ $ –  4 : (3k  – 3) = 4 : 9

If px4 + qx3 + rx2 + sx + t = (A) 33

2;x

 7 = 0 in the ratio

x;3

x%4

x ; 3  then t is equal to 3x

(B) 20

px4 + qx3 + rx2 + sx + t =

 –

x ;1 x % 3

x %1

x 2 % 3x Sol.

(D*)

1  (x  – 2) 3

x 2 % 3x 41.

(C) 1

3y  – 6 = x  – 2 x  – 3y + 4 = 0

Its y-intercept = 40.

 = 0

0

A line passes through the point (2, 2) and is perpendicular to the line 3x + y = 3. Its y-intercept is(A)

Sol.

1

[Applying R3 G R3  – R2, R1 G R1  – R2] a  – b + c  – b = 0or 2b = a + c a, b, c are in A.P.

$ $ 39.

c  – b

13 1

(C) 15

(D*) 21

x  – 1 x % 3

x %1

2 – x

x  – 3

x  – 3

x%4

3x

Putting x = 0 0 t=

1  –

3

 –

1

2 4

3  –

3  = 21

0

42.

If the system of linear equations x + 2ay + az = 0, x + 3by + bz = 0, and x + 4 cy + cz = 0 has a non-zero solution, then a, b, c (A) are in AP (B) are in GP (C*) are in HP (D) satisfy a + 2b + 3c = 0 Page # 8

Sol.

The system of linear equations has a non-zero solution, then 1 2a a

Sol.

The system of linear equations has a non-zero solution, then 1 2a a 1 3b b 1 4c

c

 = 0

Applying R2 G R2  – R1, R3 G R3  – R1 1

2a

a

0 3b – 2a b – a = 0 0 4c  – 2a c  – a

$ $ $ $

(3b  – 2a) (c  – a)  – (4c  – 2a) (b  – a) = 0 3bc  – 3ba  – 2ac + 2a2 = 4bc  – 2ab  – 4ac + 2a2 4ac  – 2ac = 4bc  – 2ab  – 3bc + 3ab 2ac = bc + ab

On dividing by abc, we get 2 1 1  =  + b a c

Hence, a, b, c are in HP. log a p 1 43.

If a, b, c are pth, qth and rth, terms of a G.P., then

(A*) 0 Sol.

(B) 1

log b q 1  equals log c r 1 (C) log abc

(D) pqr

If A be the first term and R be the c.r. of G.P., then a = ARp 1, b=ARq 1, c=ARr  –

 –

1

 –

log a = logA + (p  – 1)log R (p ; 1) log R p 1

log A p 1

& E =

log A

q 1

log A

r

1

 +

( q ; 1) log R q 1 (r ; 1) log R

r

1

p ;1 p ;1 1 = 0 + log R

q ;1 q ;1 1 r ;1 r ;1 1

 = 0

[by C2  – C1]

1 44.

log x y log x z

For positive numbers x, y, z, the numerical value of the determinant log y x

1

log z x logz y (A*) 0

(B) 1 1

Sol.

(C) 2 logx y log x z

Value of determinant log y x

1

log y z

log z x logz y 45.

1

log x log y log z 1 1 1 . . log x log y log z  = 0 = log x log y log z log x log y log z

r2

(B*) 2

1

(D) None of these

The number of values of ' r ' satisfying the equation, 39 C3r ;1; 39C (A) 1

log y z  is

(C) 3

 = 39 Cr 2 ;1; 39C 3r  is (D) 4 Page # 9

Sol.

39

C 3r ; 1 ; 39 C r 2 0 39 C r 2 ;1 ; 39 C 3r $

40

C2r  =

40

Cr 2

39

C3r ;1 % 39C3r  =

39

C r 2 ;1 % 39 Cr 2

Sol.

39

C 3r ; 1 ; 39 C r 2 0 39 C r 2 ;1 ; 39 C 3r $

40

C2r  =

40

r2 = 3r or

39

C3r ;1 % 39C3r  =

39

C r 2 ;1 % 39 Cr 2

Cr 2

r = 0, 3

or r 2 + 3r = 40 $ r = 5,  –8

Page # 10 BITSAT(XII)_PT-2_Pg.No # 1

BITSAT – XII/XIII PT – 02

BITSAT(XII)_PT-2_Pg.No # 1

BITSAT – XII/XIII PT – 02 1.

The average velocity of molecules of a gas of molecular weight M at temperature T is: (A*) 0

(B)

8 RT

3 RT M

(C)

!M

2R T

(D)

M

Sol.

Average velocity of a molecule at any temperature is zero because of its random mo tion.

2.

The ratio of r.m.s. speed to the r.ms. angular speed of a diatomic gas at certain temperature is: (assume m = mass of one molecule, M = molecular mass, "  = moment of inertia of the molecules) 3

(A)

Sol.

3I

(B)

2

3I

 (C*)

2M

2m

(D) 1

1 3 mV 2 # kT 2 2 1 2 2 "$ # kT 2 2 V # $

3" 2m

3.

A gas mixture consists of 2 moles of oxygen and 4 moles of argon at temperature T. Neglecting all vibrational modes, the total internal energy of the system is: (A) 4 R T (B) 5 R T (C) 15 R T (D*) 11 R T

Sol.

In an ideas gas internal energy = U=

4.

Sol.

f nRT 2

5 3 T. × RT + 4 × RT = 11 RT. 2 2

Maxwell ’s velocity distribution curve is given for the same quantity at two different temperatures. For  the given curves.

(A) T 1 > T2 (B*) T 1 < T 2 (C) T 1 % T2 Higher is the temperature greate r is the most probable veloc ity.

(D) T 1 = T 2

5.

In a process the density of a gas remains co nstant. If the temperature is doubled, then the change in the pressure will be: (A*) 100 % Increase (B) 200 % Increase (C) 50 % Decrease (D) 25 % Decrease

Sol.

We have

& =

PM RT

P1M P2M  = RT1 RT2

P1 P # 2 T1 2T1

P2 = 2P1 6.

12 gm He and 4 gm H 2   is filled in a container of volume 20 litre maintained at temperature 300 K. The pressure of the mixture is nearly : (A) 3 atm (B) 5 atm (C*) 6.25 atm (D) 12.5 atm

BITSAT(XII)_PT-2_Pg.No # 2 

Sol.

PV = n RT

. 12 ,

4 + ) 8 31 300

BITSAT(XII)_PT-2_Pg.No # 2 

Sol.

PV = n RT

. 12 4 + , / ) ( 8.31( 300 nRT P=  = -  4 2 *  = 6.25 × 10 5 Pa V 20 ( 10 '3 7.

In an experiment the speeds of any five molecules of an ideal gas are recorded. The experiment is repeated N times where N is very large. The aver age of recorded values, is : (A*)

2RT M

8RT

(B)

!M

(C)

3RT

(D)

M

RT M

Sol. When speed of 5 molec ules which are selected random ly, then the average is most likely to be equal to the most probable speed.

0 The average of these values is most likely equal to 8.

Sol. 9.

2 RT . M

P-V diagram o f a cyclic proc ess A 1 B 1 C 1 A is shown in figure. The temperature of the gas will be maximum at :

(A) A (B) B (C*) a point between A and B (D) a point between B and C Temperatur e at points A and B are equal. A to B temperature first in creases the n decrease. On an X temperature scale, water freezes at  – 125.0° X and boils at 375.0° X. On a Y temperature scale, water freezes at  – 70.0°Y and boils at  – 30.0°Y. The value of temperature on X-scale equal to the temperature of 50.0 °Y on Y-scale is :

& (A) 455.0° X Sol.

10.

Sol.

(B)  – 125.0° X

(C*) 1375.0° X

(D) 1500.0° X

X ' ( '125 ) Y ' ( '70 )  = 500 40 For Y = 50 X = 1375.0°X

The amount of heat supplied to decrease the volume of an ice water mixture by 1 cm 3 without any change in temperature, is equal to : ( &ice  = 0.9, &water = 80 cal/gm) (A) 360 cal (B) 500 cal (C*) 720 cal (D) None x gm ice convert into x gm water x 0.9  = 9 2 x =  –  x = 1 0 .9 0 .1 0  Q = 9 × 80 = 720 cal

11.

n moles of a gas filled in a container at temperature T is in equilibrium initially. If the gas is compressed slowly and isothermally to half its initial volume, the work done by the atmosphere on the piston is:

BITSAT(XII)_PT-2_Pg.No # 3 

(A*)

nRT 2

(B) '

nRT 2

(C) n R T

1+ .  ,- ! n 2 '  *) 2

  (D) '  n R T l  n 2

BITSAT(XII)_PT-2_Pg.No # 3 

nRT

(A*) Sol.

(B) '

2

nRT 2

(C) n R T

1+ .  ,- ! n 2 '  *) 2

  (D) '  n R T l  n 2

Work done by atmosphere = P atm 3V

&

V ................(i) 2 Initially gas in container is in thermodynamic equilibrium with its surroundings. Pressure inside cylinder = P atm PV = nRT

2

Patm V = nRT

=

Patm

As ;

0

nRT V= P atm

or

Putting in (1), W = 12.

nRT 2

In the figure shown the pressure of the gas in state B is:

(A)

63  P 25 0

(B*)

73  P 25 0

(C)

48  P 25 0

(D) none of these

Sol.

AN = 3v0 cos2 37º PB =

P0 .  16  + , v 0 / 3v 0 ( ) v 0 -  25 *

.  48 + ) = ,1 / -  25 * = P0(73/25)   Ans. 13.

(B)

A vessel contains an ideal monoatomi c gas which expands at constant pressure, when heat Q is given to it. Then the work done in expansion is: (A) Q

(B)

3 5

 Q

(C*)

2 5

 Q

(D)

2 3

Q

BITSAT(XII)_PT-2_Pg.No # 4 

.Sol.

For process at constant pressure Q = nCp 3T =

2 5 nR 3T and W= P3V = nR3T = Q 5 2

BITSAT(XII)_PT-2_Pg.No # 4 

.Sol.

For process at constant pressure Q = nCp 3T =

14.

A thermodynamic process of one mole ideal monoatomic gas is shown in figure. The efficiency of cyclic process ABCA will be :

(A) 25% Sol.

2 5 nR 3T and W= P3V = nR3T = Q 5 2

W =

(B) 12.5%

(C) 50%

(D*)

100 % 13

1 1 P 0V 0 = RT 0 . 2 2

Heat absorbed = Q AB + QBC = CVT 0 + CP2T 0 =

13  RT 0 2

1 P0 V0 2 13 .  13  + 0 Efficiency = 13 × 100 ," 2 P0 V0 # 2 RT0 ) -   * P0 V0 2 = 15.

Sol.

1 × 100 = 7.7 % 13

 

Ans.

1 mole of an ideal gas undergoes an isothermal expansion as energy is added to it as heat Q. Graph shows the volume V versus Q. The gas temperature is nearly equal to : (use R = 8.31 J/K.mole)

(A) 208.4 K (B) 268.2 K For isothermal process Q = nRT !n

(C*) 312.6 K

(D) 353.8 K

v2 v1

1800 = 1 × 8.3 T !n z get T = 312.6 K 16.

Curve in the figure shows an adiabatic compression of an ideal gas from 15 m 3 to 12 m 3, followed by an isothermal compression to a final volume of 3.0 m 3. There are 2.0 moles of the gas. Total heat supplied to the gas is equal to : ( !n2 = 0.693)

BITSAT(XII)_PT-2_Pg.No # 5 

p(Pa)

BITSAT(XII)_PT-2_Pg.No # 5 

p(Pa)

400

3

3

Sol.

12 15

V(m )

(A) 4521 J (B)  –4521 J (C*)  –6653 J (D)  –8476 J There is no heat transfer in adiabatic compression. In isothermal process

pwafd :)ks"e laihMu esa dksbZ m"ek LFkkukUrj.k ugha gksrk gSA lerkih izfØ;k esa Q=W

= P1V1 ln

= 400 x 12 ln 17.

1 =  –6653 J 4

V2 V1

Two bodies A and B have emissi vities 0.5 and 0.8 respec tively. At some temper atures the two bodies have maximum spectral emissive powers at wavelength 8000 Å and 4000 Å respectively. The ratio of  their emissive powers at these temperatures is: 5 5 (B) 10 (C) 128 16 Let the body have temperatures T 1 and T 2 respectively at wavelength 0 From Wien ’s displacement law 4  T = constant 2 41T1 = 42T2 or 8000 × T1 = 4000T 2

(A*) Sol.

(D) None of these

4 1 = 8000 Å and 42 = 4000 Å.

T1 1  = T2 2 Emissive power = e 5 AT4 0 Ratio of emissive powers at these temperature is or

e1T14

4

0 .5 5 . 1 + × , )  = = 4  = 0 .8 128 e 2T2 - 2 *

18.

N(< 100) molecules of a gas have velocities 1, 2, 3........ N km/s respectively. Then (A) rms speed and average speed of molecules is same. (B) ratio of rms speed to average speed is 7(2N + 1)(N + 1)/6N (C) ratio of rms speed to average speed is 7(2N + 1)(N + 1)/6 (D*) ratio of rms speed to average speed of molecules is 2 (2N / 1)

Sol.

V12 / V22 / ..........VN2

Vrms =

N

2

Vrms =

Vavg = Vrms Vavg

=

12 / 22 / .......... / N2 N

 =

6(N / 1)

N(N / 1) ( 2N / 1) 6N

(N / 1) ( 2N / 1) 6

N (N / 1) N/1 V1 / V2 / ........ / VN 1 / 2 / ........ / N  =  =  = 2N 2 N N

 =

2

(2 N / 1) 6 (N / 1)

BITSAT(XII)_PT-2_Pg.No # 6 

19.

A solid spherical black body of radius r and uniform mass distribution is in free space. It emits power ‘

P’ and its rate of colling is R then

(A) R P 8 r 2

(B*) R P 8 r

(C) R P 8 1/r 2

(D) R P 8

1

BITSAT(XII)_PT-2_Pg.No # 6 

19.

A solid spherical black body of radius r and uniform mass distribution is in free space. It emits power ‘

P’ and its rate of colling is R then

(A) R P 8 r 2 Sol.

(B*) R P 8 r

(C) R P 8 1/r 2

(D) R P 8

1 r

Rate of radiation per unit area is proportional to (T 4)

0 2

P 8 AT4 P 8 r 2.

Also

ms

dT dt

9 AT T4 0

dT  = R dt

9

1 r

(because m = (v &) 9 r3 and A 9 r2) 20.

A black body emits radiati on at the rate P when its abso lute tempera ture is T. At this temper ature the wavelength at which the radiation has maximum spectral emissive power is 40. If at another temperature 4 T : the power radiated is P : and wavelength at maximum spectral emissive power is 0  then 2 (A*) P : T:  = 32PT

Sol.

2

P8

1 T

1

44

&

2

(D) P : T:  = 4PT

P 8 T 4

P:  = 16 P.

P:  T: = 32PT

Thermal coefficient of volume expansion at constant pressure for an ideal gas sample of n moles having pressure P 0, volume V 0 and temperature T 0 is R (A) P V 0 0

Sol.

(C) P : T:  = 8PT

For a black body, wavelength for maximum intensity :

48

21.

(B) P :  T: = 16PT

(B)

P0 V0 R

1 (C*) T 0

1 (D) n T 0

[Easy] PV = nRT PdV = nRdT

;=

1 dV V dT

and

dV nR # dT P

For given temperature T 0 ,

;=

1 T 1

;= T 0

22.

A solid sphere of iron at 2 °C is lying at the bottom of a bucket ful l of water at 2°C. If the temperature of  the water is increased to 3 °C, the buoyant force on the sphere due to water will (A*) Increase (B) Be unchanged (C) Decrease (D) Increase or decrease depends upon the nume rical values of coeffici ent of expansion of water and iron.

Sol.

As the temperature of water is increased from 2 °C to 3°C the density of water increases (remember  anamolous behaviour of water), also the volume of sphere increases. Therefore bouyant force on sphere due to water shall increase.

23.

The lengths of two metallic rods at temperatures  > 8 2 > < 18 and from B to A if 20  – > < 2 2 > < 18 The terminal voltage across a battery of emf (A) 0 (C) < > Termianal potential across battery is : > –ir  1 If battery works as a source If battery works as a local >  + ir 1 > 1 Ideal battery or if i = 0.

> cannot be: (B) > > (D*) none of these is correct

In the circuit shown the readings of ammeter and voltmeter are 4A and 20V respectively. The meters are nonideal, then R is

(A) 5 = (C*) greater than 5 =

(B) less than 5 = (D) between 4= and 5=.

Sol. Effective resistance in the branch of R and voltmeter is ; 20  = 5= 4 Also in parallel effective resistance is less than the individual resistance. Value of R must be greater than 5 =. 0 Reff =

31.

For an adiabatic process graph between PV & V for a sample of ideal gas will be :

(A)

Sol.

(B*)

(C)

(D)

PV 8 T for adiabatic process, TV ; 1 = constant The maximum current in a galvanometer can be 10 mA. It ’s resistance is 10 =. To convert it into an ammeter of 1 Amp. a resistor should be connected in (A) series, 0.1 = (B*) parallel, 0.1 = (C) series, 100 = (D) parallel, 100=.  –

32.

Sol.

"G = 10 mA G = 10= S ("  – "G) = "G G

where S is shunt in parallel

" G G 10 ( 10 '3 ( 10 # S=  = 0.1= " ' "G 1 ' 10 ( 10 '3

BITSAT(XII)_PT-2_Pg.No # 9 

33.

Battery of internal resistor ' r ' and e.m.f. > is connected to a variable external resistance AB. If the sliding contact is moved from A to B, then terminal potential difference of battery will :

BITSAT(XII)_PT-2_Pg.No # 9 

33.

Sol.

34.

Battery of internal resistor ' r ' and e.m.f. > is connected to a variable external resistance AB. If the sliding contact is moved from A to B, then terminal potential difference of battery will :

(A) remain constant & is independent of value of external resistance (B*) increase continuously (C) decrease continuously (D) first increase and then will decrease. Terminal potential differ ence across battery will be = >  – ir  If resistance increases then ‘ i ’  will decrease So, potential will increase. Two cells of emf >1 and >2 (>2 < >1) are joined as shown in figure :

When a potentiometer is connected between x and y it balances for 300 cm length against >1. On connecting

>2 the same potentiometer between x and z it balances for 100 cm length against >1 and >2. Then the ratio > 1 is :

(A) Sol.

(B)

3 4

(C)

1 4

2 3

>2 2  = . >1 3

The equivalent resistance of the circuit across points A and B is equal to :

A B

15

10

10

15

30

20 30

20

Ans. Sol.

(D*)

..........(i) >1 = 300 8 ..........(ii)  – >  + >  = 100 8 2 1 where, 8 is the potential gradient

0 35.

1 3

(A) 22.5 = (C) Equivalent circuit is

(B) 25 =

(C*) 37.5 =

(D) 75 =

BITSAT(XII)_PT-2_Pg.No # 10 

10 A

15 B

BITSAT(XII)_PT-2_Pg.No # 10 

A

10

10

15

20

30

20

30

B

15

= 37.5 = 36.

In the circuit shown in figure find the current in branch AB of the circuit :

 A

 B

20 V

(A*) 5 (C) Sol.

11 3

A

(B) 0.5

A

(D) None of these

A

1.5A

Here in this circuit its equivalent resistance across battery can be given as

R eq

=

40 11

 A

=  B

Thus current through battery is

20 V 4A

1.5A

5.5A

20

I

=

40 11

 = 5.5

A.

Thus current 1.5 A (from figure) will be divided in 10 = & 5 = in inverse ratio thus

[rP] Vivah - 01 - Mujhe Haq Hai.mp3

I 5=

=

1.5 (10 15

 = 1 A

Thus current is branch AB  is = 1 + 4 = 5 A   Ans. I AB 

37.

What should be value of E for which galvanometer shows no deflection :

(A*) 10 V Sol.

(B) 5 V

(C) 15 V

(D) 20 V

E ' 10  = " 20

BITSAT(XII)_PT-2_Pg.No # 11

BITSAT(XII)_PT-2_Pg.No # 11

10 = E  –

E ' 10 ×5 20

40 = 4E  – E + 10 30 = 3E E = 10 V. 38.

Sol.

In the circuit shown in the figure, the potential difference between B and C is :

(A) 0.1 V (B) 2V Apply K.V.L. K.V.L. i.5 + 2 + i.3 + i.6 + i.2  – 4 = 10 16i = 12

(C) 0.5 V

(D*) 4.25 V

3 A 4 Potential difference between B and C is i=

 2 + 3 ×

39.

An ideal gas is taken through cyclic process as shown in the figure. The net work done by the gas is:

(A) zero 40.

3  = 4.25 V 4

(B*) PV

(C) 2 PV

(D) 3 PV

Heat energy absorbed by a system in going thro ugh a cyclic proces s is shown in the figure [ V in litres and p in kPa ] is:

BITSAT(XII)_PT-2_Pg.No # 12 

BITSAT(XII)_PT-2_Pg.No # 12 

(A) 10 7 ! J

T2 a 2

0

(B) 10 4 ! J

D !n4 A T= B ? C a @

1 / 2

(C*) 10 2 ! J

(D) 10 '7 ! J

!n2

=

Instructions (Q.1 & 2) Choose the alternative which can replace the italicized word : 1.

2.

He is a candid  politician (A*) frank (B) faithful

(C) soft spoken

(D) fearless

He was punished for shirking  his official work : (A) solving (B*) avoiding

(C) delegating

(D) postponing

Instructions (Q. 3 & 4) Choose the correct antonym : 3.

4.

Quell (A) Anger

(B) Query

(C) Suppress

(D*) Aggravate

Soporific (A) Inducing

(B) Inciting

(C) Consoling

(D*) Vigorous

Instructions(Q 5 & 6) Fill in the blanks with the most appropriate word from the given options. 5.

They have decided to meet the prime m inister in order to have their _______ heard. (A) agony (B) suffering (C) sorrow (D*) woes

6.

The pleasures of the world are _________ as they are not permanent. (A) tangible (B) existent (C) corporeal

(D*) illusory

Instructions (Q 7 to 9) Pick up the correct synonyms : 7.

8.

9.

Facsimile (A) laughter

(B) not genuine

(C) epithet

(D*) exact copy

Tenable (A) actual

(B) valuable

(C*) defensible

(D) ever-lasting

Stellar (A) glorious

(B) stolen

(C) outstanding

(D*) starry

Instructions(Q 10 to 13) Read the following passage and answer the questions carefully. The world of today has achieved much, but for all its declared love for humanity it has based itself far more on hatred and violence than on the virtues that make man human. War is the negation of truth and humanity. Sometimes, war may be unavoidable but its progeny are terrible to contemplate. Not mere killing, for man must die, but the deliberate and persistent propagation of hatred and falsehood, which gradually become the normal habits of the people. It is dangerous and harmful to be guided in our life's course by hatreds and aversions, for they are wasteful of energy and limit and twist the mind, and prevent if from perceiving the truth. 10.

11.

12.

The achievements of the world are not impressive because (A) there is nothing much to boast of

(B) they are mostly in the field of violence

(C*) its love of humanity is a pretence

(D) the world hasn't made any achievement

War is the negation of truth means (A) wars do not exist

(B) wars are evil

(C) wars kill human beings

(D*) wars spread and advertise falsehood

The world's declared love of humanity is (A) false

(B) true

(C) non-existent

(D*) not to be taken seriously

1

PT-02_XII_BITSAT_PAGE # 1 1

13.

Man should be guided by (A) scientific discoveries

(B) practical wisdom

(C*) generous human feelings

(D) materialism

Instruction (Q. 14 & 15) Select the option which has same relation as the given pair of words. 14.

15.

Heart  is related to Blood  in the same way as Lung  is related to (A*) Oxygen (B) Chest (C) Purification

(D) Air

Face  is related to Expression  in the same way as Hand  is related to (A*) Gesture (B) Work (C) Handshake

(D) Pointing

“

’

‘

‘

’

’

‘

‘

’

’

‘

’

Directions (16 to 19): Find the missing numbers/letter/terms : 16.

2.5, 3.5, 15, 72, 352, ? (A) 1785

(B*) 1885

Sol.

Even number cube + cube s digit s multiplication

17.

CK 10 OF 7 KM ? (A) 4, 6

Sol.

’

5 3 ?

JR TX PV (B) 6, 8

(C) 1925

(D) 1980

(C*) 6, 11

(D) 10, 12

’

Sum of alphabets numbers and then add their digits.

10 54 ? 18.

7

45 32

24 144 68 Sol. 19.

Sol.

20.

Sol. 21. Sol.

(A) 42 (B) 36 (C) 6 (D*) 4 (D) Half of the difference of top & bottom is the middle number in that column. At a dinner party every two guests used a bowl of rice between them, every three guests used a bowl of dal between them and every four used a bowl of meat between them. There were altogether 65 dishes. How many guests were present at the party? (A) 60 (B) 65 (C) 90 (D) None of these (A) Let the number of guests be x. Then, x x x number of bowls of rice = ; number of bowls of dal = ; number of bowls of meat = . 3 2 4 65 12 +  !  ( x x x 6x $ 4x $ 3x & " 60. % $ $ " 65 # " 65 # 13 x " 65 ! 12 # x " ) 2 3 4 12 *  13  ' If the following scrambled letters are rearranged to form the name of a city, the city so formed is famous for its : ACGHHIORRTT (A) Locks (B*) Cement Plant (C) Temples (D) Pottery (B) The city is CHITTORGARH and it is famous for Cement Plant. DRAMA is coded as 73 and STAGE as 25. How will you code ACTOR ? (A) 56 (B) 50 (C*) 75 (D) 67 (C) DRAMA = (4 + 18 + 1 + 13 + 1) = 37 STAGE = (19 + 20 + 1 + 7 + 5) = 52. ACTOR = (1 + 3 + 20 + 15 + 18) = 57

2

PT-02_XII_BITSAT_PAGE # 2 2